Cho \(a > 0,b > 0\). Rút gọn các biểu thức sau:
a) \(\left( {{a^{\frac{1}{2}}} + {b^{ - \frac{1}{2}}}} \right)\left( {{a^{\frac{1}{2}}} - {b^{ - \frac{1}{2}}}} \right)\);
b) \(\left( {{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right)\left( {{a^{\frac{2}{3}}} - {a^{\frac{1}{3}}}{b^{\frac{1}{3}}} + {b^{\frac{2}{3}}}} \right)\).
Advertisements (Quảng cáo)
Sử dụng kiến thức về phép tính lũy thừa: \({\left( {{a^\alpha }} \right)^\beta } = {a^{\alpha \beta }}\), \({a^{ - n}} = \frac{1}{{{a^n}}}\) với \(a \ne 0\)
a) \(\left( {{a^{\frac{1}{2}}} + {b^{ - \frac{1}{2}}}} \right)\left( {{a^{\frac{1}{2}}} - {b^{ - \frac{1}{2}}}} \right) = {\left( {{a^{\frac{1}{2}}}} \right)^2} - {\left( {{b^{ - \frac{1}{2}}}} \right)^2} = a - {b^{ - 1}} = a - \frac{1}{b}\);
b) \(\left( {{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right)\left( {{a^{\frac{2}{3}}} - {a^{\frac{1}{3}}}{b^{\frac{1}{3}}} + {b^{\frac{2}{3}}}} \right) = {\left( {{a^{\frac{1}{3}}}} \right)^3} + {\left( {{b^{\frac{1}{3}}}} \right)^3} = a + b\).